Question
Find the coefficient of x7 in the expansion of (2x2−x1)11.
Solution — Step by Step
For (a+b)n, the general term is Tr+1=(rn)an−rbr. Here a=2x2, b=−1/x, n=11.
Tr+1=(r11)(2x2)11−r(−x1)r=(r11)211−r(−1)rx2(11−r)−r
The exponent of x in Tr+1 is:
2(11−r)−r=22−3r
22−3r=7⟹3r=15⟹r=5
So the term containing x7 is T6.
coefficient=(511)⋅211−5⋅(−1)5=462⋅64⋅(−1)
=−29568
Final Answer: Coefficient of x7 is −29568.
Why This Works
The general term Tr+1 encapsulates every term of the expansion as a function of r. Setting the exponent of x to the desired value gives one equation in one unknown — solve for r, then plug back to read off the coefficient.
The negative sign comes from (−1)r with r=5 (odd). Always track the sign carefully — JEE MCQs often include both +29568 and −29568 in the options.
Alternative Method
Expand the term-by-term ratio Tr+2/Tr+1 and find when the exponent decreases past 7. Same answer, but more bookkeeping. The general-term method is the standard workhorse.
Forgetting to raise 2 to (11−r) — students sometimes write 211 as a constant out front. The an−r factor is term-specific. Always carry it inside the general term.
For binomials of the form (axp+bx−q)n, the exponent of x in Tr+1 is p(n−r)−qr. Memorize this template — JEE Main asks at least one such problem every year.